Let's call the amount Deon invested on the 7% account as x, and the amount he invested on the 8% account as y. Since Deon had $20,000 to invest, the sum of those amounts must add up to 20,000.
[tex]x+y=20000[/tex]The interest of a simple interest account is given by the following formula
[tex]I=P\times r\times t[/tex]Where P represents the principal(starting value), r represents the interest rate(in decimals) and t represents the amount of time.
Both accounts received interest for 1 year, therefore, t = 1. Now, the interest on each account will be different because the amount invested is different and the rate is different. For the first account, the interest is given by
[tex]I_x=x\times0.07\times1=0.07x[/tex]The interest on the second account will be
[tex]I_y=y\times0.08\times1=0.08y[/tex]And the total interest will be the sum of those values.
[tex]I=I_x+I_y=0.07x+0.08y=1420[/tex]Combining this equation with the first equation for the distinct amounts, we have a linear system.
[tex]\begin{cases}x+y=20000 \\ 0.07x+0.08y=1420\end{cases}[/tex]If we rewrite the first equation with x as a function of y
[tex]x+y=20000\Rightarrow x=20000-y[/tex]and substitute this on the second expression
[tex]\begin{gathered} 0.07x+0.08y=1420 \\ 0.07(20000-y)+0.08y=1420 \\ 1400-0.07y+0.08y=1420 \\ 1400+0.01y=1420 \end{gathered}[/tex]We have an expression only for y. Solving for y, we have
[tex]\begin{gathered} 1400+0.01y=1420 \\ 1400+0.01y-1400=1420-1400 \\ 0.01y=20 \\ y=\frac{20}{0.01} \\ y=2000 \end{gathered}[/tex]Using this y value on any of the previous expressions, we get our x value.
[tex]\begin{gathered} x+y=20000 \\ x+(2000)=20000 \\ x+2000-2000=20000-2000 \\ x=18000 \end{gathered}[/tex]$18,000 were investe in an account that paid 7% simple interest per year, and $2,000 in an account that paid 8% simple interest per year.